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Math Help - Help with real analysis/advanced calculus

  1. #1
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    Help with real analysis/advanced calculus

    Hey can anyone help me prove this problem?


    Let A and B be a subsets of R and bounded above.
    Let InfA=a and InfB=b.
    Let C= {xy ; x belongs to A, and y belongs to B}
    Prove that InfC=ab
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  2. #2
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    Re: Help with real analysis/advanced calculus

    Quote Originally Posted by salmaster View Post
    Let A and B be a subsets of R and bounded above.
    Let InfA=a and InfB=b.
    Let C= {xy ; x belongs to A, and y belongs to B}
    Prove that InfC=ab
    Did you mean that to read below?
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  3. #3
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    Re: Help with real analysis/advanced calculus

    This problem isn't true as stated.

    1) I think you're intending to assume that A and B are bounded below.

    2) Let A = {-1, 1}, B = {-1}. Then a = -1, b = -1, so ab = 1. Also C = {-1, 1}, so inf(C) = -1.
    Thus inf(C) is NOT equal to ab.
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  4. #4
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    Re: Help with real analysis/advanced calculus

    Yes I'm sorry I meant bounded below.
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  5. #5
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    Re: Help with real analysis/advanced calculus

    Quote Originally Posted by salmaster View Post
    Yes I'm sorry I meant bounded below.
    Let A=\{-1,2\}~\&~B=\{2,-3\} then C=\{-2,3,4,-6\}
    Does the statement hold in that case?
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  6. #6
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    Re: Help with real analysis/advanced calculus

    I don't think that we are supposed to prove it with specific examples of ordered pairs for A, B, and C..
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  7. #7
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    Re: Help with real analysis/advanced calculus

    Quote Originally Posted by salmaster View Post
    I don't think that we are supposed to prove it with specific examples of ordered pairs for A, B, and C..
    The point is that in reply #5 \inf(A)=-1,~\inf(B)=-3~\&~\inf(C)=-6 so that is a counterexample to what you are trying to show.
    So the statement you have is false.
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